Worksheet 4.3
Integrating Special Functions
Section 1 Exponential and Logarithmic Functions
Recall from worksheet 3.10 that the derivative of ex is ex . It then follows that the anti derivative
of ex is ex :
Z
ex dx = ex + c
In worksheet 3.10 we also discussed the derivative of ef (x) which is f 0 (x)ef (x) . It then follows
that
Z
f 0 (x)ef (x) dx = ef (x) + c
where f (x) can be any function. There are other ways of doing such integrations, one of which
is by substitution.
Example 1 : Evaluate the indefinite integral
We recognize that 3 =
form f 0 (x)ef (x) . Then
d(3x+2)
dx
Z
R
3e3x+2 dx.
so that the expression we are integrating has the
3e3x+2 dx = e3x+2 + c
Alternatively, we could do it by substitution: let u = 3x + 2. Then du = 3dx, and
Z
Z
3x+2
3e
dx = eu du = eu = e3x+2
Note that the integral of the function eax+b (where a and b are constants) is given by
Z
1
eax+b dx = eax+b + c
a
Example 2 : Find the area under the curve y = e5x between 0 and 2.
Z 2
A =
e5x dx
0
2
1 5x
=
e
5
0
1 10 1 0
=
e − e
5
5
1 10
=
(e − 1)
5
1
We used the property that for any real number x, x0 = 1.
Recall that the derivative of loge x is x1 . Then the anti derivative of x1 is loge x. Notice that
1
= x−1 , and that if we had used the rules we have developed to find the anti derivatives of
x
−1+1
0
things like xm , we would have the anti derivative of x−1 being x−1+1 = x0 which is not defined
as we can not divide by zero. So we have the special rule for the anti derivative of 1/x:
Z
1
dx = loge x + c
x
Recall that the derivative of loge f (x) is
Z
f 0 (x)
.
f (x)
Then we have
f 0 (x)
dx = loge f (x) + c
f (x)
R 5
R
Example 3 : Evaluate the indefinite integral 5x+2
dx. This has the form
so we get
Z
5
dx = loge (5x + 2) + c
5x + 2
f 0 (x)
f (x)
dx
Note that when you need to integrate a function like 1/(ax + b) (where a and b are constants),
then
Z
Z
1
1
a
1
dx =
dx = loge (ax + b) + c
ax + b
a
ax + b
a
Example 4 : Find the area under the curve f (x) = 1/(2x + 3) between 3 and 11.
Z
A =
3
=
=
=
=
=
11
1
dx
2x + 3
11
1
loge (2x + 3)
2
3
1
1
loge (2 × 11 + 3) − loge (2 × 3 + 3)
2
2
1
1
loge 25 − loge 9
2
2
1
1
loge (25) 2 − loge (9) 2
5
loge
3
2
Section 2 Integrating Trig Functions
To integrate trig functions we need to recall the derivatives of trig functions. We can then
work out the anti derivatives of cos x, sin x, and sec2 x. For more complicated integrals we need
special techniques that you will learn in first-year maths. The derivatives of the trig functions
are:
g(x) = sin(ax + b)
g 0 (x) = a cos(ax + b)
f (x) = cos(ax + b)
f 0 (x) = −a sin(ax + b)
h(x) = tan(ax + b)
h0 (x) = a sec2 (ax + b)
R
Example 1 : Evaluate the indefinite integral sin 3x dx.
Z
−1
sin 3x dx =
cos 3x + c
3
Note : A good way of checking your answers to indefinite integrals is to differentiate them.
You should recover the function that you started with.
Example 2 : Find the area under the curve y = cos x between 0 and π2 .
Z
π
2
A =
cos x dx
0
π
= sin x]02
π
= sin − sin 0
2
= 1 square units
Example 3 : Find
R
f (x) dx if f (x) = −3 sin(3x + 2).
Z
−3 sin(3x + 2) dx = cos(3x + 2) + c
3
Example 4 : What is the area under the curve y = sec2
Z
A =
=
=
=
=
=
π
2
x
2
between
x
dx
2
0
π
x 2
1
tan
1/2
2 0
i
x π2
2 tan
2 0
π
2 tan − 2 tan 0
4
2−0
2 square units
sec2
R
Example 5 : Evaluate the indefinite integral 5 sec2 5x dx.
Z
5 sec2 5x dx = tan 5x + c
4
π
2
and 0?
Exercises for Worksheet 4.3
1. (a) Find the anti derivative of
i. e−4x
√
ii. ex
7 − 6x
iii.
8 + 7x − 3x2
iv. cos 2x
v. sec2 (5x − 2)
1−x
vi.
x2
(b) Evaluate
Z 1
2
e2x dx
i.
0
Z 1
2x + 1
ii.
dx
2
−1 x + x + 1
Z π
4
iii.
sec2 x dx
0
Z π
2
sin2 x cos x dx
iv.
0
2. (a) Calculate the area under the curve y =
2
x+3
3x
(b) Calculate the area under the curve y = e
(c) The area under the curve y =
1
x
from x = 2 to x = 3.
from x = 0 to x = 3.
between x = 1 and x = b is 1 unit. What is b?
(d) Find the points of intersection of the curve y = sin x with the line y = 12 and hence
find the area between the two curves (from one intersection to the next). There are
two possible areas you can end up with; choose the one above y = 21 .
Z
x+6
4
x+6
(e) Show, by simple division, that
=1+
. Hence evaluate
dx.
x+2
x+2
x+2
5
Answers for Worksheet 4.3
Exercises 4.3
1. (a)
e−4x
+C
4
ii. 2ex/2 + C
sin 2x
2
v. 15 tan(5x − 2)
1
vi. − − log x
x
iv.
i. −
iii. log(8 + 7x − 3x2 ) + C
1
(e − 1)
2
ii. log 3
2. (a) 2 log 65
1
(b) (e9 − 1)
3
(c) e
(b)
iii. 1
1
iv.
3
√
π
(d) 3 −
3
(e) x + 4 log(x + 2) + C
i.
6